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Tetrahedral and 4-Tetrahedral Numbers

Asked by Warren Kendrick on November 4, 1997:
The tetrahedronal numbers: 1, 4, 10, 20, 35, . . . are derived by adding the triangular numbers: 1 + 3 + 6 + 10 + 15 + · · · + (n)(n+1)/2. At one time I derived a formula for the sum through the mth term of the tetrahedronal numbers, and also the sum of the numbers for a 4-tetrahedron: 1, 5, 15, 35, 70, . . . , which are of course the sum of the tetrahedronal numbers. (Tetrahedronal numbers are best described by placing larger and larger triangular layers of 3-spheres under the previous triangular layer of spheres, forming a tetrahedronal pyramid.) Can you provide these formulas? Thanks.
These numbers are all expressible as binomial coefficients. If we let  (IMAGE) denote the nth triangular number,  (IMAGE) the nth tetrahedral number, and so on, then


You can prove this by induction noting that each  (IMAGE) equals 1, that the formula is correct when k=2, and the difference  (IMAGE)

You could also prove this by a more "brute-force" approach, using the formulas for the sums of powers:




and so on. To arrive at these formulas you can again proceed by brute force (make the intelligent guess that  (IMAGE) should be a polynomial of degree n+1 and plug in enough values to solve for the coefficients), or else employ a trick like the following:

Write  (IMAGE) plus terms involving lower powers of m. When you add up the LHS as m ranges from 0 to n, you get  (IMAGE) When you add up the RHS, you get  (IMAGE) plus terms involving  (IMAGE) down through  (IMAGE) This gives you


The terms  (IMAGE) cancel, and you are left with a formula that expresses  (IMAGE) in terms of  (IMAGE) down through  (IMAGE) .

For example, here is how to use the above technique to arrive at the formula for  (IMAGE) :


and simplifying gives the above formula. (Note that the reason n+1 appears at the end of the second line is that we are adding up the first line as m ranges from 0 to n; that means the final constant term "1" is added up n+1 times.)

Now that you have formulas for S_k(n), it is easy to find formulas for the tetrahedral numbers. For example,


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